How to add sequence of square roots from square root 2 till square root 99 and how to add the sequence of their reciprocal here is the original problem


closed as off-topic by JMoravitz, mfl, max_zorn, user91500, José Carlos Santos Jan 27 at 9:55

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  • $\begingroup$ What makes you think that this will have a nice closed form? The result is certainly going to be irrational. $\endgroup$ – JMoravitz Jan 26 at 18:23
  • $\begingroup$ Do you want to find $S$ where $S=\frac 1{\sqrt {2}}+\frac 1{\sqrt {3}}+\frac 1{\sqrt {4}}+\ldots +\frac 1{\sqrt {97}}+\frac 1{\sqrt {98}}+\frac 1{\sqrt {99}}$? $\endgroup$ – Mohammad Zuhair Khan Jan 26 at 18:24
  • $\begingroup$ Yes that's exactly what I want. $\endgroup$ – Mai Nagah Jan 26 at 18:25
  • $\begingroup$ Do you also want $\sqrt {2}+\sqrt {3}+\sqrt {4}+\ldots +\sqrt {97}+\sqrt {98}+\sqrt {99}?$ $\endgroup$ – Mohammad Zuhair Khan Jan 26 at 18:27
  • $\begingroup$ You could express this as a generalized harmonic number, the first as $H_{99,-\frac{1}{2}}-1$ and the second as $H_{99,\frac{1}{2}}-1$ respectively, but that is really just rewriting what you already wrote without computing anything. If you want a computed value, you could easily get an approximation using any powerful enough calculator or computer using a simple for loop such as wolfram. $\endgroup$ – JMoravitz Jan 26 at 18:27

The original sums you asked for have no nice closed form and so a calculator is going to be the only feasible way to get a numerical result.

The sum you link to in the image is a totally different one and will have a nice result.

Notice that $\frac{1}{\sqrt{n}+\sqrt{n+1}} = \frac{1}{\sqrt{n}+\sqrt{n+1}}\cdot \frac{\sqrt{n}-\sqrt{n+1}}{\sqrt{n}-\sqrt{n+1}} = \frac{\sqrt{n}-\sqrt{n+1}}{n-(n+1)}=(\sqrt{n+1} - \sqrt{n})$

You have as a result:



$=\sqrt{100}-\sqrt{1} = 9$


By the way if any one is looking for the sum of sequence of square roots here is the answer



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